Given $f(x) = \frac{x²-2}{4x²+mx+4}$ determine the interval of m which makes f continuous for all real numbers.
I think the answer is $m \in (-8, 8)$ but I dont know how to get to it :/
Asked
Active
Viewed 1,423 times
2 Answers
1
Hint:
You're searching for $m$ such that $4x^2+mx+4\neq0$, i.e. when the determinant of that polynomial $\Delta=b^2-4ac$ is strictly negative.
Workaholic
- 6,763
-
So m² < 64 and therefore m < 8 and m>-8 – faruk Dec 06 '15 at 19:19
-
@faruk Correct. – Workaholic Dec 06 '15 at 19:21
1
In order for $f(x)$ to be continuous for all real numbers, you need to look at the interval of $m$ at which $4x^2 + mx + 4$ $\neq$ $0$. The quadratic equation has real roots if and only if its discriminant is positive or zero, i.e $\Delta$ $\geqslant 0$. Therefore, you need to find the values of $m$ for which the discriminant is greater than zero, in order to know the interval for which the roots are not real. $$b^2 -4ac > 0$$ The values which satisfy this inequality are $m < 8$. Hence, $f(x)$ is continuous for all real numbers only when $m<8$.
asdasd1234
- 63
-
$m\lt8$ isn't sufficient, for instance see what happens when $m=-8$ or $m=-16$. – Workaholic Dec 07 '15 at 12:15